题目描述：
Given two strings representing two complex numbers.
You need to return a string representing their multiplication. Note i2 = -1 according to the definition.
Example 1:
Input: “1+1i”, “1+1i”
Output: “0+2i”
Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.
Example 2:
Input: “1+-1i”, “1+-1i”
Output: “0+-2i”
Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.
Note:
- The input strings will not have extra blank.
- The input strings will be given in the form of a+bi, where the integer a and b will both belong to the range of [-100, 100]. And the output should be also in this form.
代码实现：
class Solution {
public:
string complexNumberMultiply(string a, string b) {
int a1=0;
int a2=0;
int b1=0;
int b2=0;
int indexa=0;
int indexb=0;
bool minusa1=false;
bool minusa2=false;
bool minusb1=false;
bool minusb2=false;
string ret="";
if(a[0]=='-'){
minusa1=true;
indexa++;
}
if(b[0]=='-'){
minusb1=true;
indexb++;
}
while(a[indexa]!='+'){
a1 = a1*10+(a[indexa]-'0');
indexa++;
}
while(b[indexb]!='+'){
b1 = b1*10+(b[indexb]-'0');
indexb++;
}
a1=minusa1==true?(0-a1):a1;
b1=minusb1==true?(0-b1):b1;
indexa++;
indexb++;
if(a[indexa]=='-'){
minusa2=true;
indexa++;
}
if(b[indexb]=='-'){
minusb2=true;
indexb++;
}
while(a[indexa]!='i'){
a2 = a2*10+(a[indexa]-'0');
indexa++;
}
while(b[indexb]!='i'){
b2 = b2*10+(b[indexb]-'0');
indexb++;
}
a2=minusa2==true?(0-a2):a2;
b2=minusb2==true?(0-b2):b2;
cout<<a1<<a2<<b1<<b2;
ret.append(to_string(a1*b1-a2*b2));
ret.append("+");
ret.append(to_string(a1*b2+b1*a2));
ret.append("i");
return ret;
}
};